Equivalence for a reversed Lipschitz-type condition Denote the real line by $\mathcal{R}$ and the non-negative real line by $\mathcal{R}^+$. Suppose that the function $f:\mathcal{R}\longrightarrow\mathcal{R}$ is continuous. Find necessary and sufficient conditions on $f$ for the following statement.
There exist a strictly increasing continuous function $\sigma:\mathcal{R}^+\longrightarrow\mathcal{R}$ satisfying $\sigma(0)=0$ such that 
$$|x-y|\leq\sigma(|f(x)-f(y)|),$$
for all $x,y\in\mathcal{R}$. 
 A: I don't think there's a simple equivalent condition for that. 
First of all we can rule out uniform continuity completely. First of all it's not necessary that it's uniformly continuous as $f(x)=x^3$ isn't but satisfy the statement. It's not sufficient either as $f(x)=1-e^{-x}$ is uniformly continuous, but doesn't satisfy the statement. This is because $f(x+1)-f(x) = e^{-x}(1 - e^{-1})$ and that converges to $0$ and since $\sigma$ is continuous we have that $\lim_{x\to\infty}\sigma( |f(x+1)-f(x)| ) = 0$ which contradicts the statement. 
What we can see is that a function that satisfies the statement will have to be strictly monotonous. We see this as it must be injective because if $f(x)=f(y)$, then $|x-y|\le\sigma(|f(x)-f(y)|) = \sigma(0) =0$ so then $x=y$. A injective continuous function must be strictly monotonous. Also since $\sigma$ is injective we have that $\sigma(|f(x+1)-f(x)|)>0$ that $|f(x+1)-f(x)|$ is bounded below which makes $f$ unbounded. The statement also implies that the inverse of $f$ is continuous making $f$ a homeomorphism on $\mathbb R$.
However we can see that this is not a sufficient condition as we can construct a counterexample by making $f$ having stationary points $a_n$ where it's successively more "flat". Make it for example $c_n (x-a_n)^3$ in a neighborhood of these and make it so that $c_n\to0$ then we have that $\sigma(1)<c_n$ which is impossible.
If we try to put restrictions on the derivate we would run into problem as there is not guaranteed that $f$ is derivable everywhere and definitely not guaranteed that $f$ has higher order derivates. Also we have that $f$ may have derivates of all orders being zero at some point (take for example $f(x) = e^{-1/x^2}$ except $f(0)=0$).
